OP Malhotra Class 9 Maths Solutions Chapter 20 Coordinates and Graphs of Simultaneous Linear Equations Exercise 20 (B)

 

Question 1. Fill in the blanks :
(i) The graph of \( x = 1 \) is a line parallel to the __________ axis.
(ii) The graph of \( y = 1 \) is a line parallel to the __________ axis.
(iii) The equation \( ax + by + c = 0 \) (where a and b are not both zero) is called __________ equation.
(iv) The graph of \( 2x = 1 \) is a line parallel to the __________ axis.
Answer:
(i) The graph of \( x = 1 \) is a line parallel to the y-axis.
(ii) The graph of \( y = 1 \) is a line parallel to the x-axis.
(iii) The equation \( ax + by + c = 0 \) (where a and b are not both zero) is called linear equation.
(iv) The graph of \( 2x = 1 \) is a line parallel to the y-axis. This is because \( 2x=1 \) can be rewritten as \( x = \frac{1}{2} \), which is a vertical line.
In simple words: Lines like \( x = \text{a number} \) are always straight up and down, parallel to the y-axis. Lines like \( y = \text{a number} \) are always straight across, parallel to the x-axis. When both \( x \) and \( y \) are in an equation (like \( ax+by+c=0 \)), it makes a straight line graph, which is called a linear equation.

🎯 Exam Tip: Remember that equations involving only 'x' (e.g., \( x=k \)) always represent vertical lines, and equations involving only 'y' (e.g., \( y=k \)) represent horizontal lines. They are perpendicular to each other.

 

Question 2. Graph the following equations :
(i) \( x = 3 \)
(ii) \( y = -4 \)
(iii) \( 2x = -7 \)
(iv) \( y = 2x \)
(v) \( y = -3x \)
(vi) \( y = x + 1 \)
(vii) \( 2x + y = 14 \)
(viii) \( 4x + 3y = 6 \)
(ix) \( x = 3y + 1 \)
Answer:
(i) \( x = 3 \)
This is a vertical line parallel to the y-axis, located 3 units to the right of the y-axis. We can also write it as \( x + 0y = 3 \). If we pick different values for \( y \) (like 1, 2, 3), the value of \( x \) will always be 3. So, we can plot points such as (3, 1), (3, 2), (3, 3) and connect them to form the line. This type of equation shows that \( x \) stays constant regardless of \( y \).
(ii) \( y = -4 \)
This is a horizontal line parallel to the x-axis, located 4 units below the x-axis. We can write this as \( 0x + y = -4 \). No matter what value we choose for \( x \) (like 1, 2, -1), the value of \( y \) will always be -4. We then plot these points, such as (1, -4), (2, -4), (-1, -4), and join them to draw the required line. The line indicates a constant y-coordinate for all x-values.
(iii) \( 2x = -7 \)
First, we solve for \( x \): \( 2x = -7 \implies x = \frac{-7}{2} \implies x = -3.5 \). This is a vertical line parallel to the y-axis, situated 3.5 units to the left. We can express this equation as \( 2x + 0y = -7 \). If we select any values for \( y \) (e.g., 1, 2, 3), the value of \( x \) will always be -3.5. We then plot points like \( \left(-\frac{7}{2}, 1\right) \), \( \left(-\frac{7}{2}, 2\right) \), and \( \left(-\frac{7}{2}, 3\right) \) and draw the line through them. This line passes through all points where the x-coordinate is -3.5.
(iv) \( y = 2x \)
To graph this equation, we find pairs of \( (x, y) \) values that satisfy it. We can pick some values for \( x \) and calculate the matching \( y \) values.

We plot these points (1, 2), (2, 4), and (-1, -2) on the graph and connect them with a straight line. This line passes through the origin (0,0) as \( y=2(0) \) is \( 0 \).
(v) \( y = -3x \)
We find several pairs of \( (x, y) \) values that make this equation true.We plot these points (1, -3), (0, 0), and (-1, 3) on the graph and connect them to get the straight line. This equation represents a line that passes through the origin and slopes downwards.
(vi) \( y = x + 1 \)
To plot this line, we find a few points that lie on it.After calculating the \( y \) values for chosen \( x \) values, we plot the points (1, 2), (0, 1), and (-1, 0) on the graph. Then, we connect these points with a straight line. This line has a positive slope and crosses the y-axis at \( y=1 \).
(vii) \( 2x + y = 14 \)
First, we rewrite the equation to solve for \( y \):
\( 2x + y = 14 \)
\( \implies \) \( y = 14 - 2x \)
Now, we choose some \( x \) values to find the corresponding \( y \) values.We then plot these points (5, 4), (6, 2), and (7, 0) on the coordinate plane and draw a straight line through them. This line shows how \( y \) decreases as \( x \) increases.
(viii) \( 4x + 3y = 6 \)
We rearrange the equation to solve for \( x \):
\( 4x + 3y = 6 \)
\( \implies \) \( 4x = 6 - 3y \)
\( \implies \) \( x = \frac{6 - 3y}{4} \)
Now, we choose some \( y \) values to find the corresponding \( x \) values.We plot the points (0, 2), (-3, 6), and (3, -2) on the graph. Then, we connect these points to draw the straight line. This line illustrates the relationship between \( x \) and \( y \) as defined by the equation.
(ix) \( x = 3y + 1 \)
For this equation, it's easier to pick values for \( y \) and then calculate \( x \).We plot these points (1, 0), (4, 1), and (-2, -1) on the graph and connect them with a straight line. This line shows how \( x \) changes as \( y \) changes.
In simple words: To graph an equation, first find some pairs of \( x \) and \( y \) numbers that make the equation true. You can do this by picking a value for \( x \) (or \( y \)) and then solving for the other. Put these pairs into a table. Then, draw dots for these points on a graph paper and connect the dots with a straight line. Remember that lines like \( x=\text{number} \) are always straight up, and lines like \( y=\text{number} \) are always straight across.

🎯 Exam Tip: Always make sure to calculate at least three points for any linear equation, even though two are enough to define a line. The third point helps you check if your calculations are correct, as all three must lie on the same straight line.

 

Question 3. Draw the graph of equation \( y = 3x - 4 \). Find graphically
(i) the values of \( y \), when \( x = -1 \).
(ii) the value of \( x \) when \( y = 5 \).
Answer:
First, we need to create a table of values for the equation \( y = 3x - 4 \). We choose some values for \( x \) and find the corresponding \( y \) values:

Next, we plot these points (0, -4), (1, -1), and (2, 2) on the graph and connect them to form the line \( y = 3x - 4 \). Once the line is drawn, we can use it to find the required values graphically:
(i) To find \( y \) when \( x = -1 \): On the graph, find \( x = -1 \). Draw a vertical line from \( x = -1 \) until it touches the graph at point P. From point P, draw a horizontal line to the y-axis. It meets the y-axis at -7. Thus, when \( x = -1 \), then \( y = -7 \).
(ii) To find \( x \) when \( y = 5 \): On the graph, find \( y = 5 \). Draw a horizontal line from \( y = 5 \) until it touches the graph at point Q. From point Q, draw a vertical line to the x-axis. It meets the x-axis at 3. Thus, when \( y = 5 \), then \( x = 3 \). Using the graph helps visualize how a change in one variable affects the other.
In simple words: First, plot the graph of the line by finding some points. Once the line is drawn, to find a \( y \) value for a given \( x \), go from the \( x \) value on the horizontal axis straight up or down until you touch the line, then go straight left or right to the \( y \) axis to read the value. To find an \( x \) value for a given \( y \), do the opposite: go from the \( y \) value on the vertical axis straight across to the line, then straight up or down to the \( x \) axis.

🎯 Exam Tip: When finding values graphically, always use dashed lines or a ruler to clearly show how you arrived at your answer from the graph. Label the points on the graph for easy identification.

 

Question 4. Find the coordinates of the point where the following lines cut the y-axis.
(i) \( y = 5x + 1 \)
(ii) \( y = 3x - 7 \)
(iii) \( y = x + 5 \)
(iv) \( 3y = 2x + 9 \)
Answer:
A line always cuts the y-axis at the point where the x-coordinate is 0. So, to find where each line cuts the y-axis, we set \( x = 0 \) in its equation and solve for \( y \). This is because the y-axis itself is the line where \( x=0 \).
(i) For the line \( y = 5x + 1 \):
Set \( x = 0 \):
\( y = 5(0) + 1 \)
\( y = 0 + 1 \)
\( y = 1 \)
So, the line cuts the y-axis at the point (0, 1).
(ii) For the line \( y = 3x - 7 \):
Set \( x = 0 \):
\( y = 3(0) - 7 \)
\( y = 0 - 7 \)
\( y = -7 \)
So, the line cuts the y-axis at the point (0, -7).
(iii) For the line \( y = x + 5 \):
Set \( x = 0 \):
\( y = 0 + 5 \)
\( y = 5 \)
So, the line cuts the y-axis at the point (0, 5).
(iv) For the line \( 3y = 2x + 9 \):
Set \( x = 0 \):
\( 3y = 2(0) + 9 \)
\( 3y = 0 + 9 \)
\( 3y = 9 \)
\( \implies \) \( y = \frac{9}{3} \)
\( \implies \) \( y = 3 \)
So, the line cuts the y-axis at the point (0, 3). For linear equations, the y-intercept can be quickly identified when the equation is in \( y = mx + c \) form, as 'c' is the y-coordinate when \( x=0 \).
In simple words: To find where any line crosses the "up and down" y-axis, you just need to put zero in place of \( x \) in the line's equation and then solve to find what \( y \) is. The point will always be (0, \( y \)).

🎯 Exam Tip: Always remember that a point on the y-axis always has an x-coordinate of zero. Substituting \( x=0 \) into the equation is the quickest way to find the y-intercept without drawing the graph.